Prove that function is continuous for all x ∈ R / {1}. f(x)=$\frac{1}{x+1}$

Question

I have to use a definition of continuity of function ( if function is continuous at point x then Δf(x) tends to 0 as Δx tends to 0) to prove than function is continuous.

This is the function:

$$\lim_{\Delta x \to 0}\frac{1}{x+1} $$ where $$x\ne1$$

For me the result is - $$-\infty$$ so the function is not continuous.

I used the formula f'(x)=$$\lim_{\Delta x \to 0}\frac{\frac{1}{x+\Delta x + 1} - \frac{1}{x+1}}{\Delta x} $$

Please assist, if I am wrong.

So, I was indeed wrong. I don't know the english word for it, but I gave to prove continuity using $$\Delta x $$

So:

$$\lim_{\Delta x \to 0}{f(x+\Delta x)-f(x)\buildrel ? \over= 0}$$

$$\lim_{\Delta x \to 0}\frac{1}{x+\Delta x +1}-\frac{1}{1+x}=\frac{-\Delta x}{(x+1)(x+1)}=-\frac{0}{(x+1)^2}\buildrel ! \over = 0$$

Is it now correct?

Answer 1

Let's show that $f$ is continuous at $x_0 \ne -1$.

For any $\varepsilon > 0$, set $\delta = \min\left\{|1+x_0|, \frac{\varepsilon|1+x_0|^2}{1+\varepsilon|1+x_0|}\right\} > 0$.

Assume $|x-x_0| < \delta$. We have $$\left|f(x) - f(x_0)\right| = \left|\frac{1}{1+x} - \frac1{1+x_0}\right| = \frac{|x_0-x|}{|1+x||1+x_0|}$$

Notice that $|1+x| \ge |1+x_0| - |x-x_0| > |1+x_0| - \delta$ so $$\left|f(x) - f(x_0)\right| < \frac{\delta}{(|1+x_0|-\delta)|1+x_0|} \le \varepsilon$$

We conclude $$(\forall \varepsilon > 0)(\exists \delta > 0)(\forall x \in \mathbb{R}\setminus \{-1\}, |x-x_0| < \delta \implies |f(x) - f(x_0)|<\varepsilon)$$ so $f$ is continuous at $x_0$.

Answer 2

The function is not defined for every $x\in\mathbb R\setminus\{1\}$ and I preassume you meant to write $\mathbb R\setminus\{-1\}$ instead of $\mathbb R\setminus\{1\}$.

Let $x_0\in\mathbb R\setminus\{-1\}$.

Then in your terms for $\Delta x\to0$ we have $f(x_0+\Delta x)=\frac1{1+x_0+\Delta x}\to\frac1{1+x_0}$, showing that $f$ is continuous at $x_0$.

This is true for every $x_0\in\mathbb R\setminus\{-1\}$, justifying the conclusion that $f$ is a continuous function.

This because $\mathbb R\setminus\{-1\}$ is the domain of the function.